What is the integral of (x^2-x+13)/((x+1)(x^2+4)) ?

The integral of (x^2-x+13)/((x+1)(x^2+4)) is 3ln|x+1|-ln|(x^2)/4+1|+1/2 arctan(x/2)+C

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integral of (x^2-x+13)/((x+1)(x^2+4))

\int \frac{x ^{2} - x + 13}{( x + 1 ) ( x ^{2} + 4 )} d x

3 ln ∣ x + 1 ∣ - ln \frac{x ^{2}}{4} + 1 + \frac{1}{2} arctan (\frac{x}{2}) + C

Solution steps

\int \frac{x ^{2} - x + 13}{( x + 1 ) ( x ^{2} + 4 )} d x

Take the partial fraction of

\frac{x ^{2} - x + 13}{( x + 1 ) ( x ^{2} + 4 )} : \frac{3}{x + 1} + \frac{- 2 x + 1}{x ^{2} + 4}

= \int \frac{3}{x + 1} + \frac{- 2 x + 1}{x ^{2} + 4} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{3}{x + 1} d x + \int \frac{- 2 x + 1}{x ^{2} + 4} d x

\int \frac{3}{x + 1} d x = 3 ln ∣ x + 1 ∣

\int \frac{- 2 x + 1}{x ^{2} + 4} d x = - ln \frac{x ^{2}}{4} + 1 + \frac{1}{2} arctan (\frac{x}{2})

= 3 ln ∣ x + 1 ∣ - ln \frac{x ^{2}}{4} + 1 + \frac{1}{2} arctan (\frac{x}{2})

Add a constant to the solution

= 3 ln ∣ x + 1 ∣ - ln \frac{x ^{2}}{4} + 1 + \frac{1}{2} arctan (\frac{x}{2}) + C

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What is the integral of (x^2-x+13)/((x+1)(x^2+4)) ?
The integral of (x^2-x+13)/((x+1)(x^2+4)) is 3ln|x+1|-ln|(x^2)/4+1|+1/2 arctan(x/2)+C